Vector diagram of loaded alternator.

Vector diagram is a part of alternator, vector diagram is a simple diagram and this diagram we represented by arrows. and The vector diagram of loaded alternator have some important symbols that is E₀ ,E ,V ,Z ,I etc. To know more about vector diagram read the article till the end.

The load on an alternator is varied its terminal voltage is also find to vary as in dc generator. This variation in terminal voltage is due to.

Voltage drop due to armature resistance R_a . Which is in phase with armature current I.
Voltage drop due to armature leakage reactance X_L .
Voltage drop due to armature reaction.

Click for what is current in electrical engineering

Armature reaction – alternator (Vector diagram of loaded alternator)

Armature reaction is the effect of armature flux on the main field flux. In case of alternator the power factor of the load has a considerable effect on the armature reaction.

So the three cases are : Unity power factor, zero power factor lagging, zero power factor leading which are explain below.

1. Unity power factor

In this case the armature flux is cross magnetising. So the result is that the flux at the leading of the pole is reduce while it is increase at the trailing tips.

So these two effect nearly offset each other leaving the average field strength constant. Hence, The armature reaction for unity power factor is distortional.

2. Zero power factor lagging

In this case the armature flux is in direct opposition to the main flux.

Hence, the main flux is decreased. And the armature reaction is demagnetising , due to which the main flux is weak and less E.M.F. is generate.

3. Zero power factor leading

So In this case armature flux will moved forward by 90^° so, that it is in phase with the main flux wave.

This result in added main flux .hence in this case ,armature reaction is magnetising ,which result in greater induced E.M.F.

4. Intermediate power factor

For intermediate power factor the effect is partly distortional and partly demagnetising because the power factor is lagging.

See also this what is a 3 phase induction motor

Synchronous impedance

For the same field excitation terminal voltage is decrease from its no-load value E_o to V. Because of

Drop due to armature resistance ( IR_a)
Drop due to leakage reactance ( IX_L )
Similarly, drop due to armature reaction.

The drop in armature voltage due to armature reaction may be accounted by the presence of reactance X_a in the armature winding. So the leakage reactance X_L and the armature reactance X_a may be combine to give synchronous reactance X_s .

Hence, X_s = X_L + X_athe total voltage drop in an alternator under load = IR_a + jIX_s = I( R_a + jX_s) = IZ_s

IR_a = Armature resistance

IX_L = Leakage reactance

X_a = Armature reactance

X_s = Synchronous impedance

E_o = Is the no-load value

V = Is the terminal voltage or rated voltage

Voltage regulation of an alternator – (Vector diagram of loaded alternator)

The voltage regulation of an alternator is defined as the rise in voltage when full-load is removed divided by the the rated terminal voltage.

So % regulation ‘up’ is

$https://latex.codecogs.com/svg.image?=\frac{E_{0}-V}V{\times 100}$

So The change in terminal voltage of an alternator depends on load on the load power factor.

NOTE : E₀ − V is the arithmetic form

For what is the relation between torque and rotor power factor

What are the methods of determining voltage regulation?

So There are different types of methods of determining voltage regulation and they are direct loading method, synchronous impedance or E.M.F. method, the ampere turn M.M.F. method,

zero power factor or potier method.

But here we consider only two method, which are explain below.

Direct loading method – (Vector diagram of loaded alternator)

The alternator is driven at synchronous speed and the terminal voltage is adjusted to its rated value ‘V’. and The load is varied until the wattmeter and ammeter indicates the rated value at desire power factor.

The entire load is thrown off while the speed and field excitation are kept constant. The open-circuit or no-load voltage E₀ reading is taken. hence, regulation can be found from % regulation

$https://latex.codecogs.com/svg.image?=\frac{E_{0}-V}V{\times 100}$

Synchronous impedance or E.M.F. method – (Vector diagram of loaded alternator)

This method requires

Armature Resistance (R_a) or stator resistance (R_a).
Open-circuit or no-load characteristic.
Short-circuit characteristic.

(i) Value of R_a

Armature resistance R_a can be measure directly by volt meter and ammeter method or by using wheatstone bridge or multimeter.

(ii) Open-circuit characteristic

In D.C. machine this is characteristic plotted by running the machine on no-load and the reading of induced voltage and field excitation current.

(iii) Short-circuit Characteristic

It is obtained by short-circuiting the armature winding through the low-resistance armature. The excitation is so adjusted to give 1.5 to 2 times the value of full load current. During this time, the speed is kept constant.

See also this what is alternator in electrical

Procedure for synchronous impedance method – (Vector diagram of loaded alternator)

open-circuit characteristic is plotted and the terminal voltage or induced voltage ‘v’ and field excitation current or no-load current I_f .
Then Short-circuit characteristic is plotted it is a straight line passing through the origin. and Bothe these curve are drawn on a common field-current base.
Consider a field current I_f the open circuit voltage corresponding to this field current is E₁ when winding is short-circuited, the terminal voltage is zero.
Hence, The whole of this voltage E₁ is being use to circulate the armature short-circuit current I₁ again the synchronous impedance Z_s.

E₁ = I₁Z_s

$https://latex.codecogs.com/svg.image?Z_{S}=\frac{E_{1}\left ( Open-circuit \right )V_{SC}}{I_{1}\left ( Short-circuit \right )I_{SC}}$

Since R_a can be found by the formula

$R_{a} = \sqrt{\left ( Z_{S} \right )^{2}-\left ( X_{S} \right )^{2}}$

5. So Knowing the value of R_a and X_s vector diagram can be draw for any load and any power factor.

OD = E₀

∴ $E_{0}= \sqrt{OB^{2}+BD^{2}}$

$E_{0}= \sqrt{\left ( VCos\Phi +IR_{a}\right )^{2}+\left ( VSin\Phi +IX_{s} \right )^{2} } Vector diagram of loaded alternator.$

So % regulation ‘up’ is =

$\frac{E_{0}-V}{V}\times 100 Vector diagram of loaded alternator.$

Some numerical and solution from Vector diagram of loaded alternator

Question-1 : A 3-phase star-connected alternator supplies a load of 10 MW at power factor 0.85 lagging and at 11 kv terminal voltage. and its resistance is 0.1 ohm per phase and synchronous reactance 0.66 ohm per phase. So calculate the line value of e.m.f. generated.

Solution :

Given data

(1) Power factor – 0.85

(2) Load – 10 MW

(3) Terminal voltage – 11 KV

(4) Resistance ( R_a ) – 0.1 ohm

(5) Synchronous resistance ( X_a ) – 0.66 ohm

Full load output current = 10 × 10⁶ / √3×11×10³×0.85 = 617.5 ≃ 618 A

IR_a = 618×0.1 = 61.8 volt

IX_s = 618×0.66 = 407.8 volt

Terminal voltage = 11,000 / √3 = 6350.8 volt

Cos Φ = 0.85

Φ = Cos⁻¹ (0.85)

Φ = 31.78

So the value of Sin Φ = 0.526

$E_{0} = \sqrt{\left ( V Cos \Phi + IR_{a} \right )^{2} + \left ( V Sin \Phi +IX_{S} \right )2}$

$E_{0} = \sqrt{\left ( 6350.8\times 0.85+61.8 \right )^{2}+ \left ( 6350.8\times 0.526 +407.8 \right )^{2}}$

E₀ = 6622.78 volt

Therefore line voltage is = √3 × 6622.7 = 11480 volt

For what is resistance ? in electrical

_{Question-2 :}_{The effective resistance of a 2200V, 50Hz, 440 KVA, 1-phase, alternator is 0.5 ohm. On short circuit, a field current of 40 ampere gives the full load current of 200 ampere. and the electromotive force on open-circuit with same field excitation is 1160 V. So Calculate the synchronous impedance and reactance.}

Solution :

Given data

(1) Power =440 KVA

(2) Voltage = 2200 V

(3) Frequency = 50Hz

(4) R_a = 0.5 ohm

(5) I.S.C. = 200 A

(6) V.O.C. =1160 V

Z_s = E₁ / I₁ = 1160/200 = 5.8 ohm

$X_{S}=\sqrt{\left ( Z_{S} \right )^{2}+\left ( R_{a} \right )^{2}} Vector diagram of loaded alternator.$

$X_{S}=\sqrt{\left ( 5.8 \right )^{2}+\left ( 0.5 \right )^{2}}$

So Synchronous reactance is = 5.77 ohm

_{Question-3 : A 60 KVA, 220V, 50Hz, 1-phase alternator has effective armature resistance of 0.016 ohm and an armature leakage reactance of 0.07 ohm Compute the voltage induced in the armature when the alternator is delivering rated current at a load current at a load power factor (i) of a unity (ii) 0.7 lagging (iii) 0.7 leading.}

Solution :

Given data

(1) Power = 60 KVA

(2) Voltage = 220 V

(3) Frequency = 50 Hz

(4) R_a = 0.016 ohm

(5) X_L = 0.07 ohm

Full load current = 60×10³ / 220 = 272.72 A

IR_a = 272.72 × 0.016 = 4.3 V

IX_L = 272.72 × 0.07 = 19 V

(i) Unity power factor

$E_{0} = \sqrt{\left ( V+IR_{a} \right )^{2}+\left ( IX_{L} \right )^{2}}$

$E_{0} = \sqrt{\left ( 220+4.3 \right )^{2}+\left ( 19 \right )^{2}}=225 volt$

(ii) Power factor – 0.7 (Lagging)

$E_{0} = \sqrt{\left ( V Cos \Phi +IR_{a} \right )^{2}+\left ( VSin \Phi + IX_{L} \right )^{2}}$

$E_{0} = \sqrt{\left ( 220\times 0.7+4.3 \right )^{2}+\left ( 220\times 0.7+19 \right )^{2}}= 234.5 volt$

So how to find out the value of Sin Φ

CosΦ = 0.7

Φ = Cos^-1 (0.7)

Φ = 45.57

So Sin Φ = 0.7

(iii) Power factor – 0.7 (Leading)

$E_{0} = \sqrt{\left ( V Cos \Phi +IR_{a} \right )^{2}+\left ( V Sin \Phi - IX_{L} \right )^{2}}$

$E_{0} = \sqrt{\left ( 220\times 0.7+4.3 \right )^{2}+\left ( 220\times 0.7-10 \right )^{2}}=208volt$

So the phasor diagram is

Question-4 : In a 50-KVA, star-connected, 440-V, 3-phase,50Hz alternator, the effective armature resistance is 0.25 ohm per phase. and The synchronous reactance is 3.2 ohm per phase and leakage reactance is 0.5 ohm per phase. Then determine at rated load and unity power factor . (i) Internal e.m.f. E_a (ii) no-load e.m.f. E_o (iii) percentage regulation on full-load (iv) value of synchronous reactance which replace armature reaction.

Solution :

Given data

(1) Power – 50 KVA

(2) Voltage – 440 V

(3) Frequency – 50 Hz

(4) R_a – 0.25 ohm

( 5) X_s – 3.2 ohm

(6) X_L – 0.5 ohm

Voltage (V) = 440 / √3 = 254.03 ≃ 254 V

Full load output current = 50×10³ / 440×√3 = 65.6 A

Voltage drop ( IR_a ) = 65.6 × 0.25 =16.4 V

Voltage drop across synchronous reactance ( IX_s ) = 65.6×3.2 = 209 V

And voltage drop across leakage reactance ( IX_L ) = 65.6×0.5 = 32.8 V

(i) $E_{a} = \sqrt{\left ( V Cos \Phi +IR_{a} \right )^{2}+\left ( IX_{L} \right )^{2}}$

$E_{a} = \sqrt{\left ( 254\times 1 +16.4\right )^{2}+\left ( 32.8 \right )^{2}}=272 volt$

Hence line value is = √3×272 = 471 volt

(ii) $E_{0}= \sqrt{\left ( V Cos \Phi +IR_{a} \right )^{2}+\left ( IX_{L} \right )^{2}}$

$E_{0}= \sqrt{\left ( 254\times 1+16.4 \right )^{2}+\left ( 209 \right )^{2}}=341.7volt$

So the Line value is = √3×341.7 = 591.8 volt

(iii) % regulation =

$\frac{E_{0}-V}{V}\times 100$

$=\frac{341.7-254}{254}\times 100$

= 34.52 %

(iv) X_a = X_s – X_L = 3.2 – 0.5 = 2.7 ohm

So Phasor diagram is

In the above article we study about vector diagram of loaded alternator so friends, what we learned from this article? How to make vector diagram and phasor diagram, how to make it in 3 different case (1st unity 2nd lagging and 3rd leading). And also we learned some important formula. Hope you like this article and know more information then please comment. share to your friends, Thank you.

Armature reaction – alternator (Vector diagram of loaded alternator)

1. Unity power factor

2. Zero power factor lagging

3. Zero power factor leading

4. Intermediate power factor

Synchronous impedance

Voltage regulation of an alternator – (Vector diagram of loaded alternator)

What are the methods of determining voltage regulation?

Direct loading method – (Vector diagram of loaded alternator)

Synchronous impedance or E.M.F. method – (Vector diagram of loaded alternator)

(i) Value of Ra

(ii) Open-circuit characteristic

(iii) Short-circuit Characteristic

Procedure for synchronous impedance method – (Vector diagram of loaded alternator)

Some numerical and solution from Vector diagram of loaded alternator

Question-1 : A 3-phase star-connected alternator supplies a load of 10 MW at power factor 0.85 lagging and at 11 kv terminal voltage. and its resistance is 0.1 ohm per phase and synchronous reactance 0.66 ohm per phase. So calculate the line value of e.m.f. generated.

(iii) Power factor – 0.7 (Leading)

Leave a ReplyCancel reply

(i) Value of R_a